Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Metric Spaces and Compactness

  1. Sep 28, 2010 #1
    My professor in lecture yesterday said that if a set is closed and bounded in a metric space it doesn't necessarily imply that it is compact. If X = R^n, then it does happen to be true, however. I was trying to construct an example, but I am getting confused. If I let X = R, and Y = (0,1) where Y is a subspace of X, then A = (0,1/2] is closed and bounded in Y. However, from where do I choose the open cover? That is, open relative to X or Y? I know in this case it won't make a difference, but maybe in differently chosen X, Y and A it might. I guess this is a matter of definition, but would like some help.

    Thanks a lot.
  2. jcsd
  3. Sep 28, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Don't think about a subset of Rn. Metric spaces just are, they aren't normally viewed as being in a larger metric space even if that is useful. In this case you have X=(0,1) with the Euclidean metric, and A=(0,1/2]. The fact that you can embed this metric space into the real numbers is irrelevant
  4. Sep 28, 2010 #3
    Ahh I see. Thanks!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook